Let $$S = \{(x,y) \in \mathbb{N} \times \mathbb{N} : 9(x-3)^2 + 16(y-4)^2 \leq 144\}$$ and $$T = \{(x,y) \in \mathbb{R} \times \mathbb{R} : (x-7)^2 + (y-4)^2 \leq 36\}$$. Then $$n(S \cap T)$$ is equal to _____

We have $$S = \{(x,y) \in \mathbb{N} \times \mathbb{N} : 9(x-3)^2 + 16(y-4)^2 \leq 144\}$$, which represents the natural number lattice points inside or on the ellipse $$\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9} \leq 1$$ centered at $$(3,4)$$ with semi-major axis $$a = 4$$ (horizontal) and semi-minor axis $$b = 3$$ (vertical). The set $$T = \{(x,y) \in \mathbb{R} \times \mathbb{R} : (x-7)^2 + (y-4)^2 \leq 36\}$$ is the disk centered at $$(7,4)$$ with radius 6.

For the ellipse, $$x$$ ranges over $$[-1, 7]$$ and $$y$$ over $$[1, 7]$$. Since we need natural numbers ($$\mathbb{N} = \{1, 2, 3, \ldots\}$$), we consider $$x \in \{1, 2, 3, 4, 5, 6, 7\}$$.

We enumerate the lattice points in the ellipse at each $$x$$-value and check which of them also satisfy the circle condition $$(x-7)^2 + (y-4)^2 \leq 36$$.

At $$x = 1$$: The ellipse gives $$16(y-4)^2 \leq 144 - 36 = 108$$, so $$(y-4)^2 \leq 6.75$$, meaning $$y \in \{2, 3, 4, 5, 6\}$$ (5 lattice points). The circle requires $$(1-7)^2 + (y-4)^2 = 36 + (y-4)^2 \leq 36$$, which forces $$y = 4$$. This gives 1 point in $$S \cap T$$.

At $$x = 2$$: The ellipse gives $$16(y-4)^2 \leq 144 - 9 = 135$$, so $$(y-4)^2 \leq 8.4375$$, meaning $$|y - 4| \leq 2$$ (since $$3^2 = 9 > 8.4375$$). So $$y \in \{2, 3, 4, 5, 6\}$$ (5 lattice points). The circle gives $$(2-7)^2 + (y-4)^2 = 25 + (y-4)^2 \leq 36$$, so $$(y-4)^2 \leq 11$$. All 5 points satisfy this. This gives 5 points.

At $$x = 3$$: The ellipse gives $$(y-4)^2 \leq 9$$, so $$|y-4| \leq 3$$, meaning $$y \in \{1, 2, 3, 4, 5, 6, 7\}$$ (7 points). The circle gives $$16 + (y-4)^2 \leq 36$$, so $$(y-4)^2 \leq 20$$. All 7 qualify. This gives 7 points.

At $$x = 4$$: Same ellipse condition as $$x = 2$$ (since $$(4-3)^2 = (2-3)^2 = 1$$): $$y \in \{2, 3, 4, 5, 6\}$$ (5 points). The circle gives $$9 + (y-4)^2 \leq 36$$, so $$(y-4)^2 \leq 27$$. All 5 qualify. This gives 5 points.

At $$x = 5$$: The ellipse gives $$16(y-4)^2 \leq 144 - 36 = 108$$, so $$(y-4)^2 \leq 6.75$$, meaning $$y \in \{2, 3, 4, 5, 6\}$$ (5 points). The circle gives $$4 + (y-4)^2 \leq 36$$. All 5 qualify. This gives 5 points.

At $$x = 6$$: The ellipse gives $$16(y-4)^2 \leq 144 - 81 = 63$$, so $$(y-4)^2 \leq 3.9375$$, meaning $$|y-4| \leq 1$$, so $$y \in \{3, 4, 5\}$$ (3 points). The circle gives $$1 + (y-4)^2 \leq 36$$. All 3 qualify. This gives 3 points.

At $$x = 7$$: The ellipse gives $$16(y-4)^2 \leq 144 - 144 = 0$$, so $$y = 4$$ (1 point). The circle gives $$0 + 0 \leq 36$$. This gives 1 point.

The total count is $$1 + 5 + 7 + 5 + 5 + 3 + 1 = 27$$.

Hence, the correct answer is $$\boxed{27}$$.